Question

To find where f(x)=ln(3+x2) is concave up, you need to determine where its second derivative f′′(x) is positive. You'll need to compute f′′(x) and then indicate the intervals where it is greater than zero.

Answer 1

To find where f(x)=ln(3+x^2) is concave up, we need to determine where its second derivative f''(x) is positive. By taking the second derivative and solving the inequality, we find that f(x) is concave up on the interval -sqrt(3) < x < sqrt(3).

Step-by-step explanation:

To find where f(x)=ln(3+x^2) is concave up, we need to determine where its second derivative f''(x) is positive. Let's start by finding the first derivative:

f'(x) = (1/(3+x^2))*2x = 2x/(3+x^2)

Now, let's find the second derivative:

f''(x) = ((2)(3+x^2) - (2x)(2x))/(3+x^2)^2

Simplifying further:

f''(x) = (6 + 2x^2 - 4x^2)/(3+x^2)^2 = (6 - 2x^2)/(3+x^2)^2

To determine where f''(x) is greater than zero, we set it to be greater than zero and solve:
(6 - 2x^2)/(3+x^2)^2 > 0

Since the numerator is always positive, the inequality reduces to:
6 - 2x^2 > 0

Simplifying further:
-2x^2 > -6
x^2 < 3
-sqrt(3) < x < sqrt(3)

Therefore, f(x)=ln(3+x^2) is concave up on the interval -sqrt(3) < x < sqrt(3).